Graded transistor labs: saturate a BJT switch, measure hFE, pick a logic-level MOSFET for 3.3 V, bias a common-emitter stage and measure its gain.
On the syllabus
BJT regions: cut-off, active and saturation
Current gain hFE and the base resistor of a switch
MOSFET threshold and logic-level gate drive
Voltage-divider bias of a common-emitter stage
Voltage gain of a common-emitter amplifier with emitter degeneration
You will be able to
Size a base resistor so a BJT switch saturates from a logic pin.
Measure a transistor’s current gain in the active region.
Choose a MOSFET that switches fully from a 3.3 V microcontroller.
Bias a common-emitter stage to the middle of its supply.
Predict and measure the gain of a common-emitter amplifier.
The first challenge is free. The rest, the viva and the certificate are part of Pro, and everything is here to read.
Graded challenges
Values like 4.7k, 100n or 2.2u are fine. Readings are the number only, in the unit shown. Every check measures the simulation, not a bench: ideal parts within their models, no noise.
1. Saturate a BJT switch from a logic pin
A saturated BJT is a closed switch: only a tenth or two of a volt from collector to emitter. To be sure it saturates, give the base several times more current than IC / hFE.
FormulaIB = (Vpin − VBE) / RB, with IB about IC / 20 to IC / 10
Worked exampleA 480 Ω relay coil on 24 V takes 50 mA. Give the base about 5 mA from a 3.3 V pin: RB = (3.3 − 0.7) / 0.005 = 520 Ω, so 510 Ω.
On the scope or meterVCE drops to a tenth or two of a volt, and the load gets almost the whole supply.
Your task
A 2N2222 switches a 120 Ω load (a small lamp, say) from 12 V. Its base is driven from a 5 V logic pin through RB.
Choose RB so the transistor is saturated, with at most 0.3 V from collector to emitter, while the pin supplies no more than 10 mA.
Rule of thumb: in saturation, give the base about a tenth of the collector current.
2. Measure hFE in the active region
In the active region the collector current is the base current multiplied by the current gain, hFE. The base resistor from the supply sets the base current.
FormulaIB = (V − VBE) / RB, hFE = IC / IB
Worked example9 V through 1 MΩ: IB = 8.3 / 1 000 000 = 8.3 µA. If IC reads 1.5 mA, hFE = 1.5 / 0.0083 = 181.
On the scope or meterMicroamps in the base and milliamps in the collector: a ratio in the hundreds.
Your task
A 2N2222 has 1 kΩ in its collector and 470 kΩ from 10 V to its base, so it sits in the active region, neither off nor saturated.
Measure the base current (through RB) in microamps and the collector current in milliamps. Their ratio is the transistor’s current gain, hFE.
3. Pick a MOSFET a 3.3 V pin can switch
A MOSFET turns on with gate voltage. The threshold is only where it starts to conduct; its low RDS(on) is reached only at the gate voltage the datasheet rates it at. Logic-level parts are rated at 4.5 V or lower.
FormulaVDS = ID × RDS(on), P = ID² × RDS(on)
Worked example3 A through 50 mΩ: VDS = 0.15 V and 0.45 W. The same 3 A in a part only half on, at 0.5 Ω, drops 1.5 V and burns 4.5 W.
On the scope or meterVDS near zero when the MOSFET is fully on. With the wrong part VDS is large and the load gets far less current.
Your task
A 3.3 V microcontroller pin drives a MOSFET’s gate through 100 Ω. The MOSFET switches a 6 Ω load (2 A) from 12 V.
Choose the MOSFET so it is fully on: at most 0.2 V from drain to source, and at least 1.9 A in the load.
Read the gate threshold and the RDS(on) rating in each datasheet: at which gate voltage is RDS(on) given?
4. Bias a common-emitter stage
With divider bias, the base voltage is set by R1 and R2. The emitter sits one VBE below it, the emitter resistor then sets the current, and that current sets the collector voltage.
FormulaIC ≈ (VB − 0.65) / RE, VC = VCC − IC × RC
Worked example9 V with RC = 3.3 kΩ and RE = 680 Ω, for VC = 4.5 V: IC = 4.5 / 3300 = 1.36 mA, VE = 0.93 V, VB = 1.58 V.
On the scope or meterThree DC voltages: base, emitter and collector. A good bias puts the collector near half the supply.
Your task
A 2N3904 has RC = 4.7 kΩ and RE = 1 kΩ on a 12 V supply. Its base is biased by a divider, R1 on top and R2 = 10 kΩ below.
Choose R1 so the collector sits near the middle of the supply, 6 V within 10%, which leaves the most room to swing both ways.
Work backwards: the collector current, then the emitter voltage, then the base voltage, then the divider.
5. Measure the gain of a common-emitter amplifier
A small signal on the base moves the emitter by the same amount. That changes the emitter current by Δv / RE, and the same current change across RC is the output. So the gain is about RC / RE, and inverted.
FormulaAv ≈ −RC / RE
Worked exampleRC = 10 kΩ and RE = 2.2 kΩ: a gain of about −4.5, so 200 mV peak to peak in gives about 900 mV out.
On the scope or meterOn the scope, the collector sine is larger than the input and upside down relative to it.
Your task
The biased stage from the last challenge now has a 1 kHz sine of 100 mV peak to peak coupled into its base through a 1 µF capacitor.
Put the scope on the collector. Measure the output swing, peak to peak, in millivolts. Then work out the voltage gain and compare it with RC / RE.
Viva questions
The questions an examiner asks at the bench. 75% passes; every answer comes with why.